sin(7π／6-α)=1/3,则cos（2π/3+2α）等于

sin(7π／6-α)=1/3,则cos（2π/3+2α）等于
sin(7π／6-α)=1/3,则cos（2π/3+2α）等于

sin(7π／6-α)=1/3,则cos（2π/3+2α）等于
sin(7π／6-α)
=cos[π/2-(7π/6-α)]
=cos(-2π/3+α)
=cos(2π/3-α)
=-cos[π-(2π/3-α)]
=-cos(π/3+α)
=1/3
cos(π/3+α)=-1/3
cos(2π/3+2α)
=cos[2(π/3+α)]
=2[cos(π/3+α)]^2-1
=-7/9

已知sin(7π／6-α)=1/3，通过合角公式可以求出sinα和cosα，然后同样通过合角公式就能得出cos（2π/3+2α）啦，具体的查查三角函数的公式表就可以啦

sin(7π／6-α)=1/3
sin(π+π/6-a)=1/3
-sin(π/6-a)=1/3
-cos(π/2-π/6+a)=1/3
cos(π/3+a)=-1/3
cos(2π/3+2a)=2cos^2(π/3+a)-1=2/9-1=-7/9.